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Question Number 82582 by M±th+et£s last updated on 22/Feb/20

solve  x^3 −3x+1=0

$${solve} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$

Commented by mr W last updated on 23/Feb/20

x^3 +3(−1)x+2((1/2))=0  p=−1  q=(1/2)  Δ=p^3 +q^2 =−1+(1/4)=−(3/4)<0  ⇒three real roots  x=2(√(−p)) sin ((1/3)sin^(−1) (q/(√(−p^3 )))+((2kπ)/3))  =2 sin ((1/3)sin^(−1) (1/2)+((2kπ)/3))  =2 sin ((π/(18))+((2kπ)/3))  (k=0,1,2)  x_1 =2 sin ((π/(18)))=0.347296  x_2 =2 sin ((π/(18))+((2π)/3))=1.532089  x_3 =2 sin ((π/(18))+((4π)/3))=−1.879385

$${x}^{\mathrm{3}} +\mathrm{3}\left(−\mathrm{1}\right){x}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${p}=−\mathrm{1} \\ $$$${q}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Delta={p}^{\mathrm{3}} +{q}^{\mathrm{2}} =−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}=−\frac{\mathrm{3}}{\mathrm{4}}<\mathrm{0} \\ $$$$\Rightarrow{three}\:{real}\:{roots} \\ $$$${x}=\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\sqrt{−{p}^{\mathrm{3}} }}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{18}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$${x}_{\mathrm{1}} =\mathrm{2}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{18}}\right)=\mathrm{0}.\mathrm{347296} \\ $$$${x}_{\mathrm{2}} =\mathrm{2}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{18}}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=\mathrm{1}.\mathrm{532089} \\ $$$${x}_{\mathrm{3}} =\mathrm{2}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{18}}+\frac{\mathrm{4}\pi}{\mathrm{3}}\right)=−\mathrm{1}.\mathrm{879385} \\ $$

Commented by M±th+et£s last updated on 23/Feb/20

thank you sir nice solution

$${thank}\:{you}\:{sir}\:{nice}\:{solution} \\ $$

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